It is known that 10% of the calculators shipped from a particular factory are defective. what is the probability that at least one in a random sample of four calculators is defective

P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)

The brobability of a binomial distribution is given by

where: n = 4

Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439

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step-by-step explanation:

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